Chapter 2 an Introduction to Electric Circuits

CHAPTER 2 AN INTRODUCTION TO ELECTRIC CIRCUITS

EXERCISE 5, Page 12

1. In what time would a current of 10 A transfer a charge of 50 C?

Charge, Q = I t from which, time, t = = 5 s

2. A current of 6 A flows for 10 minutes. What charge is transferred?

Charge, Q = I t = 6  (10  60) = 3600 C

3. How long must a current of 100 mA flow so as to transfer a charge of 80 C?

Q = It hence, time, t = = 800 s = min = 13.33 min = 13 min 20 s

EXERCISE 6, Page 15

1. The current flowing through a heating element is 5 A when a p.d. of 35 V is applied across it. Find the

resistance of the element.

Resistance of element, R = = 7

2. A 60 W electric light bulb is connected to a 240 V supply. Determine (a) the current flowing in

the bulb and (b) the resistance of the bulb.

(a) Power, P = V  I, hence, current, I = = 0.25 A

(b) Resistance, R = = 960 

3. Graphs of current against voltage for two resistors P and Q are shown inFigure 2.9. Determine the

value of each resistor.

For resistor P, resistance, R = = 2m

For resistor Q, resistance, R = = 5m

4. Determine the p.d. which must be applied to a 5 k resistor such that a current of 6 mA may

flow.

P.d., V = I  R = = 30 V

5. A 20 V source of e.m.f. is connected across a circuit having a resistance of 400 . Calculate the

current flowing.

Current, I = = 0.05 A or 50 mA

EXERCISE 7, Page 17

1. The hot resistance of a 250 V filament lamp is 625 . Determine the current taken by the lamp

and its power rating.

Current, I = = 0.4 A

Power rating, P = V  I = 250  0.4 = 100 W

(orP = = 100 W or P = = 100 W)

2. Determine the resistance of a coil connected to a 150 V supply when acurrent of (a) 75 mA

(b) 300 μA flows through it.

(a) Resistance, R = = 2k

(b) Resistance, R = 500 k or 0.5 MΩ

3. Determine the resistance of an electric fire which takes a current of 12 A from a 240 V supply.

Find also the power rating of the fire and the energy used in 20 h.

Resistance, R = = 20 

Power rating, P =V  I = 240  12 = 2880 W or 2.88 kW

Energy = power  time = 2.88 kW  20 h = 57.6 kWh

4. Determine the power dissipated when a current of 10 mA flows through an appliance having a

resistance of 8 k.

Power, P = = 0.8 W

5. 85.5 J of energy are converted into heat in 9 s. What power is dissipated?

Power, P = = 9.5 W

6. A current of 4 A flows through a conductor and 10 W is dissipated. Whatp.d. exists across the ends of

the conductor ?

Power, P = V  I hence p.d., V = = 2.5 V

7. Find the power dissipated when: (a) a current of 5 mA flows through a resistance of 20 k

(b) a voltage of 400 V is applied across a 120 k resistor (c) a voltage applied to a resistor is

10 kV and the current flow is 4 mA

(a) Power, P = = 0.5 W

(b) Power,P = = 1.33 W

8. A battery of e.m.f. 15 V supplies a current of 2 A for 5 min. How much energy is supplied in this

time?

Energy = power  time = (V  I)  t = (15  2)  (5  60) = 9000 J or 9 kJ

9. A d.c. electric motor consumes 72 MJ when connected to 400 V supply for 2 h 30 min. Find the

power rating of the motor and the current taken from the supply.

Energy = power  time from which, power = = 8 kW

Power = V  I from which, current = = 20 A

10. A p.d. of 500 V is applied across the winding of an electric motor andthe resistance of the winding is

50 . Determine the power dissipated by the coil.

Power dissipated by coil,P = = 5000 W = 5 kW

11. In a household during a particular week three 2 kW fires are used on average 25 h each and

eight 100 W light bulbs are used on average 35 h each. Determine the cost of electricity for the

week if 1 unit of electricity costs 15p

Energy in week = 3(2 kW  25 h) + 8(kW  35 h) = 150 + 28 = 178 kWh

Cost = 178  15 = 2670p = £26.70

12. Calculate the power dissipated by the element of an electric fire of resistance 30  when a

current of 10 A flows in it. If the fire is on for 30 hours in a week determine the energy used.

Determine also the weekly cost of energy if electricity costs 13.50p per unit.

Power, P = = 3000 W or 3 kW

Energy = power  time = 3 kW  30 h = 90 kWh

Cost =90  13.50p = 1215p = £12.15

EXERCISE 8, Page 19

1. A television set having a power rating of 120 W and electric lawnmower of power rating 1 kW

are both connected to a 250 V supply. If 3 A, 5 A and 10 A fuses are available state which is the

most appropriate for each appliance.

Power, P = V I hence, current, I =

For the television, I = = 0.48 A, hencethe 3 A fuse is the most appropriate

For the lawnmower, I = = 4 A, hencethe 5 A fuse is the most appropriate